College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 160: 62


The solutions are $x=1\pm\dfrac{\sqrt{6}}{2}$

Work Step by Step

$2x^{2}-4x-1=0$ Take out common factor $2$ from the left side: $2\Big(x^{2}-2x-\dfrac{1}{2}\Big)=0$ Take the $2$ to divide the right side: $x^{2}-2x-\dfrac{1}{2}=\dfrac{0}{2}$ $x^{2}-2x-\dfrac{1}{2}=0$ Take $\dfrac{1}{2}$ to the right side: $x^{2}-2x=\dfrac{1}{2}$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-2$: $x^{2}-2x+\Big(\dfrac{-2}{2}\Big)^{2}=\dfrac{1}{2}+\Big(\dfrac{-2}{2}\Big)^{2}$ $x^{2}-2x+1=\dfrac{1}{2}+1$ $x^{2}-2x+1=\dfrac{3}{2}$ Factor the left side of the equation, which is a perfect square trinomial: $(x-1)^{2}=\dfrac{3}{2}$ Take the square root of both sides: $\sqrt{(x-1)^{2}}=\pm\sqrt{\dfrac{3}{2}}$ $x-1=\pm\dfrac{\sqrt{6}}{2}$ Solve for $x$: $x=1\pm\dfrac{\sqrt{6}}{2}$
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